THAM KHẢO NHA 

https://olm.vn/hoi-dap/detail/56559485893.html

Đặt \(M=\frac{1}{4}+\frac{1}{16}+.....+\frac{1}{10000}\)

\(M=\frac{1}{2.2}+\frac{1}{4.4}+\frac{1}{6.6}+......+\frac{1}{100.100}\)

\(=\frac{1}{2.2}\left(1+\frac{1}{2.2}+\frac{1}{3.3}+.....+\frac{1}{50.50}\right)\)

\(< \frac{1}{2.2}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{49.50}\right)\)

\(=\frac{1}{4}.\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\right)\)

\(=\frac{1}{4}.\left(1+1-\frac{1}{50}\right)< \frac{1}{4}.\left(1+1\right)=\frac{1}{4}.2=\frac{1}{2}\)

Vậy \(M< \frac{1}{2}\)

99/100

cho mk nhé

\(Đ\text{ặt }S=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+....+\frac{1}{10000}\)

\(S=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)

\(S=\frac{1}{2^2}\cdot\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)

Ta có :

\(\frac{1}{2^2}< \frac{1}{1\cdot2};\text{ }\frac{1}{3^2}< \frac{1}{2\cdot3};\text{ }...;\text{ }\frac{1}{50^2}< \frac{1}{49\cdot50}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+1=2\)

\(\Rightarrow\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< \frac{1}{2^2}\cdot2\)

\(\Rightarrow S< \frac{1}{2}\) (ĐPCM)

Đặt \(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+....+\frac{1}{10000}\)

\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{100^2}\)

\(\Rightarrow4A=1+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{50^2}\)

\(\Rightarrow4A< 1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)

\(\Rightarrow4A=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(\Rightarrow4A< 2-\frac{1}{50}< 2\)

\(\Rightarrow4A< 2\Rightarrow A< \frac{2}{4}=\frac{1}{2}\)

=>a<1/2

\(A=\frac{1}{4}+\frac{1}{16}+....+\frac{1}{576}\)

\(\Rightarrow4A=1+\frac{1}{4}+....+\frac{1}{144}\)

\(\Rightarrow4A-A=1-\frac{1}{576}\)

\(\Rightarrow3A=1-\frac{1}{576}\)

\(\Rightarrow A=\frac{1-\frac{1}{576}}{3}\)

Ta có: \(\frac{1-\frac{1}{576}}{3}< \frac{1}{3}< \frac{1}{2}\)

vậy suy ra đpcm

A=14 +116 +....+1576 

⇒4A=1+14 +....+1144 

⇒4A−A=1−1576 

⇒3A=1−1576 

⇒A=1−1576 3 

Ta có: 1−1576 3 <13 <12 

vậy suy ra đpcm

S=1/4+1/9+1/16+1/25+1/36+1/49+1/64+1/81=1-1/81=1/81

80/81 là đúng

vô lí vì 2/2 = 1 mà 8/9 < 1

A=\(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}\)=\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)

=>A<\(\frac{1}{2.2}+\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}+\frac{1}{12.14}\)

=>A<\(\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{12}-\frac{1}{14}\right)\)\(:2\)=\(\left(\frac{1}{2}-\frac{1}{14}\right):2\)<\(\frac{1}{2}\)

=>A<\(\frac{1}{2}\)